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# If f : [– 5, 5] R is a differentiable function and if f (x) does not vanish anywhere, then prove that f (– 5) is not equal to f (5).

Q3     If  $f ; [ -5 ,5] \rightarrow R$ is a differentiable function and if   $f ' (x)$   does not vanish
anywhere, then prove that $f (-5) \neq f(5)$

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It is given that
$f ; [ -5 ,5] \rightarrow R$ is a differentiable function
Now, f is a differential function. So, f is also a continuous function
We obtain the following results
a ) f is continuous in [-5,5]
b ) f is differentiable in (-5,5)
Then, by Mean value theorem we can say that there exist a c  in (-5,5) such that
$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$
$f^{'}(c) = \frac{f(5)-f(-5)}{5-(-5)}\\ f^{'}(c)= \frac{f(5)-f(-5)}{10}\\ 10f^{'}(c)= f(5)-f(-5)$
Now, it is given that $f ' (x)$   does not vanish anywhere
Therefore,
$10f^{'}(c)\neq 0\\ f(5)-f(-5) \neq 0\\ f(5)\neq f(-5)$
Hence proved

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